Answer
$$\frac{{25}}{6}{\sin ^{ - 1}}\left( {\frac{{3x}}{5}} \right) + \frac{{x\sqrt {25 - 9{x^2}} }}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {25 - 9{x^2}} dx} \cr
& {\text{Let 3}}x = 5\sin \theta ,{\text{ }}dx = \frac{5}{3}\cos \theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\sqrt {25 - 9{x^2}} dx} = \int {\sqrt {25 - 9{{\left( {\frac{5}{3}\sin \theta } \right)}^2}} \left( {\frac{5}{3}\cos \theta d\theta } \right)} \cr
& = \int {\sqrt {25 - 25{{\sin }^2}\theta } \left( {\frac{5}{3}\cos \theta d\theta } \right)} \cr
& = \frac{5}{3}\int {\sqrt {25 - 25{{\sin }^2}\theta } \cos \theta d\theta } \cr
& = \frac{5}{3}\int {\sqrt {25\left( {1 - {{\sin }^2}\theta } \right)} \cos \theta d\theta } \cr
& = \frac{5}{3}\int {\sqrt {25\left( {{{\cos }^2}\theta } \right)} \cos \theta d\theta } \cr
& = \frac{5}{3}\int {5\cos \theta \cos \theta d\theta } \cr
& = \frac{{25}}{3}\int {{{\cos }^2}\theta d\theta } \cr
& {\text{Use the power reducing formula co}}{{\text{s}}^2}\theta {\text{ = }}\frac{{1 + \cos 2\theta }}{2} \cr
& = \frac{{25}}{6}\int {\left( {1 + \cos 2\theta } \right)d\theta } \cr
& {\text{Integrate}} \cr
& {\text{ = }}\frac{{25}}{6}\left( {\theta + \frac{1}{2}\sin 2\theta } \right) + C \cr
& {\text{ = }}\frac{{25}}{6}\left( {\theta + \frac{1}{2}\sin \theta \cos \theta } \right) + C \cr
& {\text{We know that }}\sin \theta = \frac{{3x}}{5},{\text{ }}\cos \theta = \frac{{\sqrt {25 - 9{x^2}} }}{5} \cr
& {\text{ = }}\frac{{25}}{6}\left( {{{\sin }^{ - 1}}\left( {\frac{{3x}}{5}} \right) + \frac{1}{2}\left( {\frac{{3x}}{5}} \right)\left( {\frac{{\sqrt {25 - 9{x^2}} }}{5}} \right)} \right) + C \cr
& {\text{ = }}\frac{{25}}{6}\left( {{{\sin }^{ - 1}}\left( {\frac{{3x}}{5}} \right) + \frac{{3x\sqrt {25 - 9{x^2}} }}{{25}}} \right) + C \cr
& {\text{ = }}\frac{{25}}{6}{\sin ^{ - 1}}\left( {\frac{{3x}}{5}} \right) + \frac{{x\sqrt {25 - 9{x^2}} }}{2} + C \cr} $$
